1st box- yes or no 2nd box- only positive values or only negative values or zero

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1st box- yes or no 2nd box- only positive values or only negative values or zero Score: 0.5 of 1 pt 5 of 9 (9 complete) v HW Score: 67.22%, 6.0.. ia Question Help 9.3.13-T Proctored Nonproctored A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple u u1 H2 random samples selected from normally distributed populations, and do not n 35 31 assume that the population standard deviations are equal. Complete parts (a) X 75.55 84.46 and (b) below. Use a 0.01 significance level for both parts. s 10.99 20.44 Mc. Fai to reject Ho. There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. Fail to reject Ho. There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. b. Construct a confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. 18.843 -w2 1.023 Round to three decimal places as needed.) Does the confidence interval support the conclusion of the test? v because the confidence interval contains henk Answer.

Explanation / Answer

x1 = 75.55 , s1 = 10.99 , x2 = 84.46 , s2 = 20.44 , n1 = 35 , n2= 31

z value at 99% CI = 2.576

CI = (x1 - x2) + /- z * sqrt(s1^2/n1 + s2^2/n2)
= ( 75.55 - 84.46) + /- 2.576 * sqrt( 10.99^2/35 + 20.44^2/ 31)
= (-19.508 , 1.688)

Yes, because the confidence interval contains (-19.508 , 1.688)

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